NumPy: calculate averages with NaNs removed

Question

How can I calculate matrix mean values along a matrix, but to remove nan values from calculation? (For R people, think na.rm = TRUE).

Here

Answer 1

From numpy 1.8 (released 2013-10-30) onwards, nanmean does precisely what you need:

>>> import numpy as np
>>> np.nanmean(np.array([1.5, 3.5, np.nan]))
2.5

Answer 2

# I suggest you this way:
import numpy as np
dat  = np.array([[1, 2, 3], [4, 5, np.nan], [np.nan, 6, np.nan], [np.nan, np.nan, np.nan]])
dat2 = np.ma.masked_invalid(dat)
print np.mean(dat2, axis=1)   

Answer 3

How about using Pandas to do this:

import numpy as np
import pandas as pd
dat = np.array([[1, 2, 3], [4, 5, np.nan], [np.nan, 6, np.nan], [np.nan, np.nan, np.nan]])
print dat
print dat.mean(1)

df = pd.DataFrame(dat)
print df.mean(axis=1)

Gives:

0    2.0
1    4.5
2    6.0
3    NaN

Answer 4

Assuming you've also got SciPy installed:

http://www.scipy.org/doc/api_docs/SciPy.stats.stats.html#nanmean

Answer 5

One more speed check for all proposed approaches:

Python 2.7.11 |Anaconda 2.4.1 (64-bit)| (default, Jan 19 2016, 12:08:31) [MSC v.1500 64 bit (AMD64)]
IPython 4.0.1 -- An enhanced Interactive Python.

import numpy as np
from scipy.stats.stats import nanmean    
dat = np.random.normal(size=(1000,1000))
ii = np.ix_(np.random.randint(0,99,size=50),np.random.randint(0,99,size=50))
dat[ii] = np.nan
In[185]: def method1():
    mdat = np.ma.masked_array(dat,np.isnan(dat))
    mm = np.mean(mdat,axis=1)
    mm.filled(np.nan) 

In[190]: %timeit method1()
100 loops, best of 3: 7.09 ms per loop
In[191]: %timeit [np.mean([l for l in d if not np.isnan(l)]) for d in dat]
1 loops, best of 3: 1.04 s per loop
In[192]: %timeit np.array([r[np.isfinite(r)].mean() for r in dat])
10 loops, best of 3: 19.6 ms per loop
In[193]: %timeit np.ma.masked_invalid(dat).mean(axis=1)
100 loops, best of 3: 11.8 ms per loop
In[194]: %timeit nanmean(dat,axis=1)
100 loops, best of 3: 6.36 ms per loop
In[195]: import bottleneck as bn
In[196]: %timeit bn.nanmean(dat,axis=1)
1000 loops, best of 3: 1.05 ms per loop
In[197]: from scipy import stats
In[198]: %timeit stats.nanmean(dat)
100 loops, best of 3: 6.19 ms per loop

So the best is 'bottleneck.nanmean(dat, axis=1)' 'scipy.stats.nanmean(dat)' is not faster then numpy.nanmean(dat, axis=1).

Answer 6

'''define dataMat'''
numFeat= shape(datMat)[1]
for i in range(numFeat):
     meanVal=mean(dataMat[nonzero(~isnan(datMat[:,i].A))[0],i])